Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}4x+8y &= -8 \\ -4x+y &= -5\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $-4x = -y-5$ Divide both sides by $-4$ to isolate $x$ $x = {\dfrac{1}{4}y + \dfrac{5}{4}}$ Substitute this expression for $x$ in the first equation. $4({\dfrac{1}{4}y + \dfrac{5}{4}}) + 8y = -8$ $y + 5 + 8y = -8$ Simplify by combining terms, then solve for $y$ $9y + 5 = -8$ $9y = -13$ $y = -\dfrac{13}{9}$ Substitute $-\dfrac{13}{9}$ for $y$ in the top equation. $4x+8( -\dfrac{13}{9}) = -8$ $4x-\dfrac{104}{9} = -8$ $4x = \dfrac{32}{9}$ $x = \dfrac{8}{9}$ The solution is $\enspace x = \dfrac{8}{9}, \enspace y = -\dfrac{13}{9}$.